A Note on the Integral Transforms on a Function Space I,II

By

Bong Jin KIM

Abstract

In this paper we obtain some results of integration by parts formulas involving integral transformsof functionals of the form $F(y)=f(\langle\theta_{1},y\rangle,$

$\ldots,$$(\theta_{n},y\})$ for s-a.e. $y\in C_{0}[0, T]$ , where $\langle\theta,y\rangle$ denotes the

Riemann-Stieltjes integral $\int_{0}^{T}\theta(t)dy(t)$ . Furthermore we obtain the various relationships that exist amongthe integral transform, the convolution product and the first variation for a class of functionals definedon $K(Q)$ , the space of complex-valued continuous functions on $Q=[0,S]\cross[0,T]$ which satisfy $x(s,0)=$

$x(O,t)=0$ for all $0\leq s\leq S$ and $0\leq t\leq T$ . Also we obtain Parseval‘s and Plancherel‘s relations for theintegral transform of some functionals defined on $K(Q)$ .

\S 1. Parts formulas involving integral transforms on function space

In a unifying paper [15], Lee defined an integral transform $\mathcal{F}_{\alpha\beta}$ of analytic functionals onan abstract Wiener space. For certain values of the parameters $\alpha$ and $\beta$ and for certain classesof functionals, the Fourier-Wiener transform [3], the Fourier-Feynman transform [4] and theGauss transform are special cases of his integral transform $\mathcal{F}_{\alpha\beta}$ . In [6], Chang, Kim and Yooestablished an interesting relationship between the integral transform and the convolution prod-uct for functionals on an abstract Wiener space. In this paper we establish several integrationby parts formulas involving integral transforms, convolution products, and the first variations offunctionals of the form $F(y)=f(\{\theta_{1},y\rangle,$

$\ldots,$$\langle\theta_{n},y\rangle)$ for s-a.e. $y\in C_{0}[0, T]$ , where $(\theta,y\rangle$ denotes

the Riemann-Stieltjes integral $\int_{0}^{T}\theta(t)dy(t)$ .Let $C_{0}[0, T]$ denote one-parameter Wiener space; that is the space of all real-valued contin-

uous functions $x(t)$ on $[0,T]$ with $x(O)=0$ . Let $\mathcal{M}$ denote the class of all Wiener measurablesubsets of $C_{0}[0,T]$ and let $m$ denote Wiener measure. $(C_{0}[0, T],\mathcal{M},m)$ is a complete measurespace and we denote the Wiener integral of a Wiener integrable functional $F$ by

(1.1) $\int_{C_{0}[0,T]}F(x)m(dx)$ .

2010 Mathematics Subject Classification(s): $28C20$.Key Words: integral transform, convolution product, first variation, integration by parts formula, Wiener integral,Yeh-Wiener integral, Parseval’s relation, Plancherel‘s relation

数理解析研究所講究録第 1797巻 2012年 20-41 20

Let $\alpha$ and $\beta$ be nonzero complex numbers. Next we state the definitions of the integraltransform $\mathcal{F}_{\alpha\beta}F$ , the convolution product $(F*G)_{\alpha}$ and the first variation $\delta F$ for functionalsdefined on $K=K[0, T]$ , the space of complex-valued continuous functions defined on $[0, T]$

which vanish at $t=0$.

Definition 1.1. Let $F$ be a functional defined on $K$. Then the integral transform $\mathcal{F}_{\alpha,\beta}F$ of $F$

is defined by

(1.2) $\mathcal{F}_{\alpha\beta}(F)(y)\equiv \mathcal{F}_{\alpha_{:}\beta}F(y)\equiv\int_{C_{0}[0,T]}F(\alpha x+\beta y)m(dx)$ , $y\in K$

if it exists [6, 12, 13, 15].

Definition 1.2. Let $F$ and $G$ be functionals defined on $K$. Then the convolution product$(F*G)_{(\chi}$ of $F$ and $G$ is defined by

(1.3) $(F*G)_{\alpha}(y) \equiv\int_{C_{0}[0,T]}F(\frac{y+\alpha x}{\sqrt{2}})G(\frac{y-\alpha x}{\sqrt{2}})m(dx)$, $y\in K$

if it exists[6, 10, 12, 19, 21].

Definition 1.3. Let $F$ be a functional defined on $K$ and let $w\in K$ . Then the first variation$\delta F$ of $F$ is defined by

(1.4) $\delta F(y|w)\equiv\frac{\partial}{\partial t}F(y+tw)|_{t=0}$, $y\in K$

if it exists [2, 5, 12, 17].

Let $\{\theta_{1},\theta_{2}, \ldots\}$ be a complete orthonormal set of real-valued functions in $L_{2}[0, T]$ and as-sume that each $\theta_{j}$ is of bounded variation on $[0, T]$ . Then for each $y\in K$ and $j\in\{1,2, \ldots\}$ , theRiemann-Stieltjes integral $\langle\theta_{j},y\}\equiv\int_{0}^{T}\theta_{j}(t)dy(t)$ exists. Furthermore

(1.5) $| \langle\theta_{j},y\}|=|\theta_{j}(T)y(T)-\int_{0}^{T}y(t)d\theta_{j}(t)|\leq C_{j}\Vert y\Vert_{\infty}$

with

(1.6) $C_{j}=|\theta_{j}(T)|+Var(\theta_{j}, [0, T])$ ,

where $Var(\theta_{j}, [0, T])$ denote the total variation of $\theta_{j}$ on $[0, T]$ .Next we describe the class of functionals which is related to this paper. For $0\leq\sigma<1$ , let

$E_{\sigma}$ be the space of all functionals $F$ : $Karrow \mathbb{C}$ of the form

(1.7) $F(y)=f(\{\vec{\theta},y\})=f((\theta_{1},y\rangle, \ldots, \langle\theta_{n},y\})$

21

for some positive integer $n$ , where $f(\vec{\lambda})=f(\lambda_{1}, \ldots,\lambda_{n})$ is an entire function of the $n$ complexvariables $\lambda_{1},$ $\ldots,\lambda_{n}$ of exponential type; that is to say,

(1.8) $|f(\vec{\lambda})|\leq A_{F}\exp\{B_{F}|\vec{\lambda}|^{1+\sigma}\}$

for some po$:;itive$ constants $A_{F}$ and $B_{F}$ , where $| \vec{\lambda}|^{1+\sigma}=\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}$ .

In addition we use the notation

$F_{j}(y)=f_{j}(\{\vec{\theta},y\rangle)$

where $f_{j}(/1)arrow=\tau_{\lambda_{j}}^{\partial_{-f(/l_{1},\ldots,l_{n})}}$’ for $j=1,$ $\ldots,n$ .Recently [12], Kim, Kim and Skoug established the results that if $F$ and $G$ are elements

of $E_{\sigma}$ then.$F_{\alpha\beta}F,$ $(F*G)_{a},$ $\delta F(\cdot|w)$ and $\delta F(y|\cdot)$ are also elements of $E_{\sigma}$ and examined var-ious relationships holding among $\mathcal{F}_{a\beta}F,$ $\mathcal{F}_{a,\beta}G,$ $(F*G)_{a},$ $\delta F$ and $\delta G$ . For related work see[3, 6, 10, 12, 15, 17, 19, 21] and for a detailed survey of previous work see [18].

We introduce the following three existence theorems for the integral transform, the convo-lution $produ|.t$ and the first variation of functionals in $E_{\sigma}[12]$ .

Theorenn 1.4. Let $F\in E_{\sigma}$ be given by (1.7). Then the integml transform $\mathcal{F}_{\alpha\beta}F$ exists,

belongs to $E_{\sigma}$ . and is given by the formula

(1.9) $\mathcal{F}_{a\beta}F(y)=h(\{\vec{\theta},y\})$

for $y\in K,$ $w/^{1}lere$

(1.10) $h( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\alpha\vec{u}+\beta\vec{\lambda})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$

where $\Vert\vec{u}\Vert^{2}:=\sum_{j=1}^{n}u_{J^{2}}$ and $d\vec{u}=du_{1}\cdots du_{n}$ .

Theoren11.5. Let $F,G\in E_{\sigma}$ be given by (1.7) with corresponding entirefunctions $f$ and $g$,

respectively. Then the convolution $(F*G)_{\alpha}$ exists, belongs to $E_{\sigma}$ and is given by the fomula

(1.11) $(F*G)_{\alpha}(y)=k(\langle\vec{\theta},y\rangle)$

for $y\in K,$ wjlere

(1.12) $k( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\frac{\vec{\lambda}+\alpha\tilde{u}}{\sqrt{2}})g(\frac{\vec{\lambda}-\alpha\vec{u}}{\sqrt{2}})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$.

Theorem 1.6. Let $F\in E_{\sigma}$ be given by (1.7) and let $w\in K$ . Then

(1.13) $\delta F(y|w)=p(\{\vec{\theta},y\})$

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for $y\in K$, where

(1.14) $p( \vec{\lambda})=\sum_{j=1}^{n}\{\theta_{j},w\}f_{j}(\vec{\lambda})$ .

Furthermore, as afimction of$y\in K,$ $\delta F(y|w)$ is an element of $E_{\sigma}$ .

Now we state some observations which we use later in this paper. First of all, equation (1.2)

implies that

(1.15) $\mathcal{F}_{\alpha\beta}F(y/\sqrt{2})=\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)$

for all $y\in K$ . Next, a direct calculation using (1.4), (1.2), (1.13) and (1.15) shows that

$\delta \mathcal{F}_{\alpha,\beta}F(y/\sqrt{2}|w/\sqrt{2})=\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}F(y|w)$

(1.16)$= \frac{\beta}{\sqrt{2}}\sum_{j=1}^{n}\{\theta_{j},w\rangle \mathcal{F}_{\alpha\beta/\sqrt{2}}F_{j}(y)$

for all $y$ and $w$ in $K$ . Finally, by similar calculations, we obtain that

(1.17) $\mathcal{F}_{\alpha\beta}(\delta F(\cdot|w))(y/\sqrt{2})=\frac{\sqrt{2}}{\beta}\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}F(y|w)$

for all $y$ and $w$ in $K$, and for all $y\in K$,

(1.18) $(\mathcal{F}_{\alpha\beta}F)_{j}(y)=\beta \mathcal{F}_{\alpha,\beta}F_{j}(y)$ .

Let

$A=$ {$y\in C_{0}[0,$ $T]:y$ is absolutely continuous on $[0,$ $T]$ with $y’\in L^{2}[0,$ $T]$ }.

We note that if we choose $z\in L_{2}[0, T]$ and define $w(t)= \int_{0}^{t}z(s)ds$ for $t\in[0, T]$ , then $w$ is anelement of $A,$ $w’=z$ a.e. on $[0, T]$ , and for all $v\in L_{2}[0, T],$ $\{v, w\}=(v,w’)=(v,z)$ , where$(v,z)= \int_{0}^{T}v(s)z(s)ds$ .

The following theorem plays a key role throughout this paper. In this theorem the Wienerintegral of the first variation of functional $F$ is expressed in terms of the Wiener integral of $F$

multiplied by a linear factor.

Theorem 1.7. Let $F\in E_{\sigma}$ be given by (1.7) and $w\in A$ , then

(1.19) $\int_{C_{0}[0,T]}\delta F(x|w)m(dx)=\int_{C_{0}[0,T]}F(x)\{z,x\rangle m(dx)$

where $w(t)= \int_{0}^{t}z(s)ds$ on $[0, T]$ for some $z\in L_{2}[0, T]$ .

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Proof. Let $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L^{2}[0,T]$ . Using the Gram-Schmit process we canfind an orthonormal set $\{\theta_{1}, \ldots,\theta_{n},\theta_{n+1}\}$ with $\theta_{n+1}=\frac{1}{\Vert z_{n+1}||}z_{n+1}$ , where

$z_{n+1}=z- \sum_{j=1}^{n}(\theta_{j},z)\theta_{j}$ .

Then by the Wiener integration formula

$\int_{C_{0}[0,T]}F(x)\{z,x\}m(dx)$

$= \int_{C_{0}[0,7]}f(\langle\vec{\theta},x\})(\sum_{j=1}^{n}(\theta_{j},z)\langle\theta_{j},x\}+\Vert z_{n+1}\Vert\{\theta_{n+1},x\})m(dx)$

$=(2 \pi)^{-(\mathfrak{l}2+1)/2}\int_{\mathbb{R}^{n+1}}f(u\gamma(\sum_{j=1}^{n}(\theta_{j},z)u_{j}+\Vert z_{n+1}\Vert u_{n+1})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}-\frac{1}{2}u_{n+1}^{2}\}du_{n+1}d\vec{u}$.

If we evaluate the last integral with respect to $u_{n+1}$ , we obtain

$\int_{C_{0}[0,T]}F(x)\{z,x\}m(dx)=(2\pi)^{-n/2}\sum_{j=1}^{n}(\theta_{j},z)\int_{\mathbb{R}^{n}}f(u\gamma u_{j}\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$.

On the other hand, since $w\in A\subset K$ , by Theorem 1.6

$\delta F(x|w)=\sum_{j=1}^{n}\{\theta_{j},w\}f_{j}(\{\vec{\theta},x\})=\sum_{j=1}^{n}(\theta_{j},z)f_{j}(\{\vec{\theta},x\})$.

Hence by $th|^{\sim}$ Wiener integration formula

$(_{C_{0}[0,T]} \delta F(x|w)m(dx)=\sum_{j=1}^{n}(\theta_{j},z)\int_{q)[0,T]}f_{j}(\{\vec{\theta},x\})m(dx)$

$=(2 \pi)^{-n/2}\sum_{j=1}^{n}(\theta_{j},z)\int_{\mathbb{R}^{n}}f_{j}(\vec{u})\exp\{-\frac{1}{2}\Vert u\Vert^{2}\}d\vec{u}$.

Note that for each $j=1,$ $\ldots,n$, the integration by parts formula yields

$\int_{\mathbb{R}}f_{j}(u\gamma\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}$

$= \lim_{abarrow\infty}\lim_{arrow-\infty}[f(\vec{u})\exp\{-\frac{1}{2}u_{j}^{2}\}]_{a}^{b}+\int_{\mathbb{R}}f(\vec{u})u_{j}\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}$.

But since $f1\lfloor S$ of exponential type, the double limit in the last equation is equal to $0$ and so

$\int_{\mathbb{R}}f_{j}(\hat{u})\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}=\int_{\mathbb{R}}f(u\gamma u_{j}\exp\{-\frac{1}{2}u_{j}^{2}\}du_{j}$ .

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Hence

$\int_{C_{0}[0,T]}\delta F(x|w)m(dx)=(2\pi)^{-n/2}\sum_{j=1}^{n}(\theta_{j},z)\int_{\mathbb{R}^{n}}f(u\gamma u_{j}\exp\{-\frac{1}{2}\Vert u\Vert^{2}\}d\vec{u}$

and this completes the proof. $\square$

In our next theorem we obtain an integration by parts formula for the products of functionalsin $E_{\sigma}$ .

Theorem 1.8. Let $F,$ $G\in E_{\sigma}$ be given by (1.7) with corresponding entire fiunctions $f$ and$g$, respectively. Thenfor $w\in A$ , we have the following integmtion by parts formula.

(1.20) $\int_{C_{0}[0,T]}[F(y)\delta G(y|w)+\delta F(y|w)G(y)]m(dy)=\int_{C_{0}[0,T]}F(y)G(y)(z,y\rangle m(dy)$ ,

where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ .

Proof. Define $H(y)=F(y)G(y)$ for $y\in K$ and let $h(/t)arrow=f(\vec{A})g(/t)arrow$ . Then $H\in E_{\sigma}$ and

$\delta H(y|w)=\sum_{j=1}^{n}\{\theta_{j},w\rangle f_{j}(\{\vec{\theta},y\})g(\{\vec{\theta},y\rangle)+f(\langle\vec{\theta},y\})\sum_{j=1}^{n}\{\theta_{j},w\rangle g_{j}(\{\vec{\theta},y\})$

$=\delta F(y|w)G(y)+F(y)\delta G(y|w)$ .

Thus equation (1.20) follows from Theorem 1.7. $\square$

By choosing $G=F$ in Theorem 1.8, we obtain the following corollary.

Corollary 1.9. Let $F\in E_{\sigma}$ be given by (1.7). Thenfor each $w\in A$ ,

(1.21) $\int_{C_{0}[0,T]}F(y)\delta F(y|w)m(dy)=\frac{1}{2}\int_{C_{0}[0,T]}[F(y)]^{2}\{z,y\}m(dy)$,

where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ .

As we saw in Theorem 1.6 above if $F$ belongs to $E_{\sigma}$ , then $\delta F(y|w_{1})$ also belongs to $E_{\sigma}$

as a function of $y$ . Thus if we replace $G(y)$ with $\delta F(y|w1)$ in Theorem 1.8, then we have thefollowing corollary.

Corollary 1.10. Let $F\in E_{\sigma}$ be given by (1.7). Then for each $w_{1},w2\in A$,

$\int C_{0}[0,T]^{[F(y)\delta^{2}F(\cdot|w_{1})(y|w_{2})+\delta F(y|w)\delta F(y|w_{1})]m(dy)}2$

(1.22)

$= \int C_{0}[0,T]^{F(y)\delta F(y|w)\langle Z2,\mathcal{Y}\rangle m(dy)}l$

where $w_{i}(t)= \int_{0}^{t}z_{i}(s)ds$for some $z\iota\in L_{2}[0, T],$ $i=1,2$.

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As we saw in Theorem 1.4 above if $G$ belongs to $E_{\sigma}$ , then $\mathcal{F}_{a\beta}G$ also belongs to $E_{\sigma}$ . Thusif we replace $G$ with $\mathcal{F}_{a}pG$ in Theorem 1.8, then we have the following corollary.

Corollary 1.11. Let $F,G\in E_{\sigma}$ be given as in Theorem 1.8. Then for each $w\in A$ ,

$\int_{C_{0}[0,T]}[F(y)\delta \mathcal{F}_{a\beta}G(y|w)+\delta F(y|w)\mathcal{F}_{\alpha\beta}G(y)]m(dy)$

(1.23)

$= \int_{C_{0}[0,T]}F(y)\mathcal{F}_{a}pG(y)\langle z,y)m(dy)$ ,

where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0,T]$ .

By replacing $F$ and $G$ by $\mathcal{F}_{a,\beta}F$ and $\mathcal{F}_{a\beta}G$ , respectively, in Theorem 1.8, we obtain thefollowing corollary.

Corollary 1.12. Let $F,G\in E_{\sigma}$ be as in Theorem 2.5. Then for each $w\in A$ ,

$\int_{ci)[0,T]}[\mathcal{F}_{\alpha\beta}F(y)\delta \mathcal{F}_{a\beta}G(y|w)+\delta \mathcal{F}_{a\beta}F(y|w)\mathcal{F}_{\alpha\beta}G(y)]m(dy)$

(1.24)$= \int_{cb[0,T]}\mathcal{F}_{a\beta}F(y)\mathcal{F}_{a\beta}G(y)(z,y\rangle m(dy)$,

where $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ .

\S 2. Various integration formulas and examples

In this section we establish various integration formulas involving integral transforms, con-volution products and first variations. Furthermore we give some examples to illustrate theintegration formulas in this paper.

In [12], Kim, Kim and Skoug established various relationships holding among $\mathcal{F}_{\alpha_{I}\beta}F,$ $\mathcal{F}_{a\beta}G$,

$(F*G)_{a},$ $\delta F$ and $\delta G$ . From these relationships and the results in Section 1 above, we can estab-lish various integration formulas.

From Theorem 1.7 above we know that the Wiener integral of the first variation of functional$F\in E_{\sigma}$ is expressed in terms of the Wiener integral of $F$ multiplied by a linear factor. On theother hand, some of the formulas, for example, Formulas 3.3, 3.5, 4.1, 4.2, and 5.2 in [12] giveus the expressions of the first variation of various functionals. Hence it is easy to obtain thefollowing formulas (2.1) through (2.7) below. We just state the formulas without proofs.

Let $w\in A$ with $w(t)= \int_{0}^{t}z(s)ds$ for some $z\in L_{2}[0, T]$ throughout this section. The paper[12] was concemed with the class $E_{0}$ . Bm as commented in Remark 5.6 of that paper, all theformulas in [12] still true for functionals in $E_{\sigma}$ . Hence we will assume that $F\in E_{\sigma}$ in Formula2.1 through Formula 2.6 and Corollary 2.7 below.

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Formula 2.1. From Formula 3.3 of [7], we have

(2.1) $\beta\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta}\delta F(\cdot|w)(y)m(dy)=\int_{C_{0}[0,T]}\mathcal{F}_{a\beta}F(y)\{z,y\}m(dy)$ .

Formula 2.2. From Formula 3.5 of [12], we have

$\sum_{j=1}^{n}\frac{\langle\theta_{j},w\rangle}{\sqrt{2}}l_{C_{0}[0,T]}[(F_{j}*G)_{\alpha}(y)+(F*G_{j})_{\alpha}(y)]m(dy)$

(2.2)

$= \int_{C_{0}[0,T]}(F*G)_{(\chi}(y)\{z,y\rangle m(dy)$

and if $F=G$,

(2.3) $\sqrt{2}\sum_{j=1}^{n}\{\theta_{j}, w\}\int_{C_{0}[0,T]}(F*F_{j})_{\alpha}(y)m(dy)=\int_{C_{0}[0,T]}(F*F)_{\alpha}(y)\{z,y\}m(dy)$.

Formula 2.3. From Formula 4.1 of [12], we have

$\int_{C_{0}[0,T]}(\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}G(y|w)+\delta \mathcal{F}_{\alpha\beta/\sqrt{2}}F(y|w)\mathcal{F}_{\alpha\beta/\sqrt{2}}G(y))m(dy)$

(2.4) $= \int_{C_{0}[0,T]}(\mathcal{F}_{\alpha,\beta}F(\frac{y}{\sqrt{2}})\delta \mathcal{F}_{\alpha\beta}G(\frac{y}{\sqrt{2}}|\frac{w}{\sqrt{2}})+\delta \mathcal{F}_{r\beta}F(\frac{y}{\sqrt{2}}|\frac{w}{\sqrt{2}})\mathcal{F}_{cx\beta}G(\frac{y}{\sqrt{2}}))m(dy)$

$= \int_{C_{0}[0,T]}\mathcal{F}_{\alpha/3}(F*G)_{\alpha}(y)\langle z,y\rangle m(dy)$

and if $F=G$,

$\int_{C_{0}[0,T]}\mathcal{F}_{\alpha,\beta}[F(\frac{y}{\sqrt{2}})][\delta \mathcal{F}_{\alpha\beta}F(\frac{y}{\sqrt{2}}|\frac{w}{\sqrt{2}})]m(dy)$

(2.5)

$= \frac{1}{2}\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta}(F*F)_{\alpha}(y)\{z,y\}m(dy)$ .

Formula 2.4. From Formula 4.2 of [12], we have

$\frac{\beta}{\sqrt{2}}\sum_{j=1}^{n}\{\theta_{j},w\}\int_{C_{0}[0,T]}[(\mathcal{F}_{a\beta}F_{j}*\mathcal{F}_{\alpha\beta}G)_{\alpha}(y)+(\mathcal{F}_{\alpha\beta}F*\mathcal{F}_{\alpha\beta}G_{j})_{a}(y)]m(dy)$

(2.6)

$= \int_{C_{0}[0,T]}(\mathcal{F}_{\alpha\beta}F*\mathcal{F}_{\alpha\beta}G)_{\alpha}(y)\{z,y\}m(dy)$ .

Formula 2.5. From Formula 5.2 of [12] we have,

$\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta}(\delta F(\cdot|w)G(\cdot)+F(\cdot)\delta G(\cdot|w))(\frac{y}{\sqrt{2}})m(dy)$

(2.7)$= \frac{\sqrt{2}}{\beta}\int_{C_{0}[0,T]}(\mathcal{F}_{(x\beta}F*\mathcal{F}_{\alpha\beta}G)_{i\alpha/\beta}(y)\langle z,y\}m(dy)$.

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Using the equations (1.10) through (1.12) we obtain the following integration formula forthe Wiener integral of the integral transform with respect to the first argument of the variation.

Formula 2.6. For $F\in E_{\sigma}$ we have

(2.8) $\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}(\delta F(\cdot|w))(y)m(dy)=\sum_{j=1}^{n}\langle\theta_{j},w\}\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}F_{j}(y)m(dy)$.

Proof. By (1.16) and Theorem 1.7 we have

$\frac{\beta}{\sqrt{2}}\sum_{j=1}^{n}\langle\theta_{j},w\}\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}F_{j}(y)m(dy)=\int_{C_{0}[0,T]}\mathcal{F}_{a\beta/\sqrt{2}}F(y)\{z,y\rangle m(dy)$ .

Similarly by (1.17) and Theorem 1.7 we have

$\frac{\beta}{\sqrt{2}}\int_{C_{0}[0,T]}\mathcal{F}_{\alpha\beta/\sqrt{2}}(\delta F(\cdot|w))(\frac{y}{\sqrt{2}})m(dy)=\int_{C_{0}[0,T]}\mathcal{F}_{l\beta/\sqrt{2}}F(y)\langle z,y\}m(dy)$ .

Thus we have the above formula (2.8). $\square$

We next obtain an integration formula for functionals which is a product of elements of $E_{\sigma}$

by some linear factors.

Corollary 2.7. Let $k$ be a natural number and let $z_{j}\in L_{2}[0, T]$ for $j=1,2,$ $\ldots,k+1$ . Let$F\in E_{\sigma}$ and let

$F^{[k]}(y)=F^{[k-1]}(y) \langle z_{k},y\rangle=F(y)\prod_{j=1}^{k}\langle Zj,\mathcal{Y}\rangle$ .

Then we have the following integml equation.

$\int_{C_{0}[0,T]}F^{[k+1]}(y)m(dy)$

(2.9)$= \int_{q)[0,T]^{\delta F^{[k-1]}(y|w_{k+1})\{Zk,\mathcal{Y}\}m(d_{\mathcal{Y}})+\{Zk}},w_{k+1}\rangle\int_{C_{0}[0,T]}F^{[k-1]}(y)m(dy)$

where $F^{[0]}=F$ and $w_{k+1}(t)= \int_{0}^{t}z_{k+1}(s)ds$ .

Proof. To prove this theorem we simply take the first variation of the $F^{[k]}(y)=F^{[k-1]}(y)\{z_{k},y\}$ .Now we have

$\delta F^{[k]}(y|w_{k+1})=\frac{\partial}{\partial t}(F^{[k-1]}(y+tw_{k+1})\langle zk,\mathcal{Y}+tw_{k+1}\})|_{t=0}$

$=\delta F^{[k-1]}(y)\{z_{k},y\rangle+F^{[k-1]}(y)\{z_{k},w_{k+1}\rangle$ .

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Hence we have

$\int_{C_{0}[0,T]}\delta F^{[k]}(y|w_{k+1})m(dy)$

$= \int c_{0[0,T]^{\delta F^{[k-1]}(y)\langle z_{k},y\rangle m(dy)+\{w_{k+1}\}\int_{C_{0}[0,T]}F^{[k-1]}(y)m(dy)}}Zk$,

But by Theorem 1.7,

$\int_{C_{0}[0,T]}\delta F^{[k]}(y|w_{k+1})m(dy)=\int_{C_{0}[0,T]}\delta F^{[k]}\{z_{k+1},y\}m(dy)=\int_{C_{0}[0,T]}F^{[k+1]}(y)m(dy)$

and this completes the proof. $\square$

We finish this section by giving some examples for the illustration of the integration by partsformulas.

Example 2.8. Let $F(y)= \sum_{j=1}^{n}\langle\theta_{j},y\}$ which is an element of $E_{\sigma}$ , then we have $\delta F(y|w)=$

$F(w)$ , where $w(t)= \int_{0}^{t}z(s)ds\in A$ . Thus we can obtain

$\int_{C_{0}[0,T]}\delta F(y|w)m(dy)=\int_{C_{0}[0,T]}F(w)m(dy)=\sum_{j=1}^{n}\{\theta_{j},w\rangle=\sum_{j=1}^{n}\int_{0}^{T}\theta_{j}(s)z(s)ds$.

Since the constant functional $G\equiv 1$ belongs to $E_{\sigma}$ and its first variation equals to zero, Theorem1.8 yields the following.

(2.10) $\sum_{j=1}^{n}\int_{C_{0}[0,T]}\{\theta_{j},y\rangle\langle z,y\}m(dy)=\sum_{j=1}^{n}\int_{0}^{T}\theta_{j}(s)z(s)ds$.

Example 2.9. Let $G(y)= \exp\{\sum_{j=1}^{n}\{\theta_{j},y\}\}$ which is an element of $E_{\sigma}$ , then we have

$\delta G(y|w)=\sum_{j=1}^{n}\{\theta_{j},w\}\exp\{\sum_{j=1}^{n}(\theta_{j},y\}\}=\sum_{j=1}^{n}\{\theta_{j},$ $w\rangle G(y)$ ,

where $w(t)= \int_{0}^{t}z(s)ds\in A$ . Hence by the Wiener integration formula

$\int_{C_{0}[0,T]}\delta G(y|w)m(dy)=\sum_{j=1}^{n}\{\theta_{j},w\rangle\int_{C_{0}[0,T]}\sum_{j=1}^{n}\exp\{\langle\theta_{j},y\}\}m(dy)$

$=e^{n/2} \sum_{j=1}^{n}(\theta_{j},w\}$ .

From Theorem 1.8 we obtain the following Wiener integral.

(2.11) $\int_{C_{0}[0,T]}\exp\{\sum_{j=1}^{n}\{\theta_{j},y\rangle\}\{z,y\rangle m(dy)=e^{n/2}\sum_{j=1}^{n}\{\theta_{j},w\rangle$.

29

Example 2.10. Let $H(y)= \sum_{j=1}^{n}[\{\theta_{j},y\rangle]^{2}$ which is an element of $E_{\sigma}$ , then we have,

$\delta H(y|w)=2\sum_{j=1}^{n}\langle\theta_{j},w\rangle(\theta_{j},y\}=2\sum_{j=1}^{n}(\theta_{j},z)\langle\theta_{j},y\}$ .

where $w(t)= \int_{0}^{t}z(s)ds\in A$ . Thus we can obtain the following by Wiener integration formula.

$\int_{C_{0}[0,T]}\delta H(y|w)m(dy)=2\sum_{/=1}^{n}(\theta_{j},z)\int_{C_{0}[0,T]}\{\theta_{j},y\rangle m(dy)=0$.

By Theorem 1.8 we have the following.

(2.12) $\sum_{j=1}^{n}\int_{C_{0}[0,T]}[\langle\theta_{j},y\rangle]^{2}\{z,y\rangle m(dy)=0$ .

Example 2.11. Let $L(y)=[ \sum_{j=1}^{n}\langle\theta_{j},y\rangle]^{2}$ which is an element of $E_{\sigma}$ , then we have,

$\delta \mathcal{F}_{\alpha\beta}L(y|w)=2\beta^{2}\sum_{j=1}^{n}(\theta_{j},w\rangle\sum_{j=1}^{n}\langle\theta_{j},y\rangle$ .

From the Wiener integration formula, we have the followings.

$\int_{C_{0}[0,T]}\delta \mathcal{F}_{a\beta}L(y|w)m(dy)=\psi^{2}\sum_{j=1}^{n}(\theta_{j},w\rangle\int_{C_{0}[0,T]}\sum_{j=1}^{n}\langle\theta_{j},y\rangle m(dy)=0$

and

$\int_{C_{0}[0,T]}\mathcal{F}_{a\beta}L(y)\langle z,y\}m(dy)$

$= \int_{C_{0}[0,T]}[n\alpha^{2}+|\beta\sum_{j=1}^{n}\{\theta_{j},y\rangle]^{2}]\{z,y\rangle m(dy)$

$=0+ \beta^{2}\int_{C_{0}[0,T]}[\sum_{j=1}^{n}(\theta_{j},y\rangle]^{2}\langle z,y\rangle m(dy)$.

From the Corollary 1.11 we have the following.

(2.13) $\int_{C_{0}[0,T]}[\sum_{j=1}^{n}\langle\theta_{j},y\}]^{2}\{z,y\rangle m(dy)=0$ .

30

Examples 2.8 through 2.11 are interesting to note that we can obtain the Wiener integrals onthe left hand side of (2.10) through (2.13) by using Theorem 1.8 or Corollary 1.11 rather thandirect calculation using Wiener integration formula.

\S 3. Integral transforms of functionals on function space of two variables

Recently [12] Kim, Kim and Skoug studied the relationships that exist among the integraltransform, the convolution product and the first variation for functionals defined on $K[0, T]$ , thespace of complex-valued continuous functions on $[0, T]$ which vanish at zero. In this paper weextend the results in [12] for functionals of two variables.

Let $Q=[0,S]\cross[0, T]$ and let $C(Q)$ denote Yeh-Wiener space; that is, the space of all real-valued continuous functions $x(s,t)$ on $Q$ with $x(s,O)=x(0,t)=0$ for all $0\leq s\leq S$ and $0\leq t\leq$

$T$ . Yeh [18] defined a Gaussian measure $m_{Y}$ on $C(Q)$ (later modified in [20]) such that as astochastic process $\{x(s,t);(s,t)\in Q\}$ has mean $E[x(s,t)]=0$ and covariance $E[x(s,t)x(u,v)]=$$\min\{s,u\}\min\{t,v\}$ .

Let $\mathcal{M}$ denote the class of all Yeh-Wiener measurable subsets of $C(Q)$ and we denote theYeh-Wiener integral of a Yeh-Wiener integrable functional $F$ by

(3.1) $\int_{C(Q)}F(x)m_{Y}(dx)$ .

Let $K(Q)$ be the space of complex-valued continuous functions defined on $Q$ and satisfying$x(s,0)=x(0,t)=0$ for all $0\leq s\leq S$ and $0\leq t\leq T$ . Let $\alpha$ and $\beta$ be nonzero complex numbers.Next we state the definitions of the integral transform $\mathcal{F}_{\alpha\beta}F$ , the convolution product $(F*G)_{\alpha}$

and the first variation $\delta F$ for functionals defined on $K(Q)$ .

Definition 3.1. Let $F$ be a functional defined on $K(Q)$ . Then the integral transform $\mathcal{F}_{\alpha\beta}F$

of $F$ is defined by

(3.2) $\mathcal{F}_{\alpha fl}F(y)=\int_{C(Q)}F(\alpha x+\beta y)m_{Y}(dx)$, $y\in K(Q)$

if it exists [6, 12, 13, 15].

It is obvious that (3.2) implies that

$\mathcal{F}_{\alpha\beta}F(\frac{y}{\sqrt{2}})=\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)$

for all $y\in K(Q)$ .

Definition 3.2. Let $F$ and $G$ be functionals defined on $K(Q)$ . Then the convolution product$(F*G)_{\alpha}$ of $F$ and $G$ is defined by

(3.3) $(F*G)_{\alpha}(y)= \int_{C(Q)}F(\frac{y+\alpha x}{\sqrt{2}})G(\frac{y-\alpha x}{\sqrt{2}})m_{Y}(dx)$, $y\in K(Q)$

if it exists [6, 10, 12, 19,21].

31

Definition 3.3. Let $F$ be a functional defined on $K(Q)$ and let $w\in K(Q)$ . Then the firstvariation $\delta F$ of $F$ is defined by

(3.4) $\delta F(y|w)=\frac{\partial}{\partial t}F(y+tw)|_{t=0}$ , $y\in K(Q)$

if it exists [2,5, 14, 16].

Now we introduce a concept of the function of bounded variation of two variables, and anintegration by parts formula for a Riemann-Stieltjes integral for functions of two variables. Theconcept of bounded variation for a function of two variables is surprisingly complex. There areseveral nonequivalent definitions. The paper [7] by Clarkson and Adams is useful in sortingout many of the relationships between the various definitions. In this paper we will use thedefinition used by Hardy and Krouse [1,11] which we now review.

Let $R=[a,b]\cross[c,d]$ and let $P$ be a partition of $R$ given by

$a=s0<s_{1}<\cdots<s_{n}=b$ , $c=t_{0}<t_{1}<\cdots<t_{m}=d$ .

A function $f(s,t)$ is said to be of bounded variation on $R$ in the sense of Hardy and Krouseprovided the following three conditions hold.(a) There is a constant $k$ such that

(3.5) $\sum_{i=1}^{n}\sum_{j=1}^{m}|f(s_{i},t_{j})-f(s_{i},t_{j-1})-f(s_{i-1},t_{j})+f(s_{i-1},t_{j-1})|\leq k$

for all partition $P$.(b) For each $t\in[c,d],$ $f(\cdot,t)$ is a function of bounded variation on $[a,b]$ .(c) For each $s\in[a,b],$ $f(s, \cdot)$ is a function of bounded variation on $[c,d]$ .

The total variation $Var(f,R)$ of $f$ over $R$ is defined to be the supremum of the sums in (3.5)

over all partitions $P$ of R. $Var(f(\cdot,t), [a,b])$ and $Var(f(s, \cdot), [c,d])$ will denote the total variationsof $f(\cdot,t)$ on $[a,b]$ and $f(s, \cdot)$ on $[c,d]$ , respectively, as functions of single variable.

The definition of bounded variation used by Hardy and Krouse has the important propertythat if $g$ is continuous on $R$ and $f$ is of bounded variation on $R$ then the Riemann-Stieltjes inte-grals $\int_{R}g(s,t)df(s,t)$ and $\int_{R}f(s,t)dg(s,t)$ both exist and are related by the following integrationby parts formula [9].

Theorem 3.4. Let $R=[a,b]\cross[c,d]$ . Let $g(s,t)$ be afunction of bounded variation in thesense ofHardy and Kmuse and let $f(s,t)$ be a continuous function on R. Then the followingintegmtion by partsformula holds.

$\int_{R}g(s,t)df(s,t)=[f(s,t)g(s,t)]_{R}-l^{d}[f(s,t)dg(s,t)]_{a}^{b}$

(3.6)

$- \int_{a}^{b}[f(s,t)dg(s,t)]_{c}^{d}+\int_{R}f(s,t)dg(s,t)$

32

where

$[f(s,t)g(s,t)]_{R}=f(b,d)g(b,d)-f(a,d)g(a,d)-f(b,c)g(b,c)+f(a,c)g(a,c)$ ,

$[f(s,t)dg(s,t)]_{a}^{b}=f(b,t)dg(b,t)-f(a,t)dg(a,t)$

for each $t\in[c,d]$ , and

$[f(s,t)dg(s,t)]_{c}^{d}=f(s,d)dg(s,d)-f(s,c)dg(s,c)$

for each $s\in[a,b]$ .

Let $\{\theta_{1},\theta_{2}, \ldots,\theta_{n}\}$ be an orthonormal set of real-valued functions in $L_{2}(Q)$ . Furthermoreassume that each $\theta_{j}$ is of bounded variation in the sense of Hardy and Krouse on $Q$ . Then foreach $y\in K(Q)$ and $j=1,2,$ $\ldots$ , the Riemann-Stieltjes integral $\langle\theta_{j},y\}\equiv\int_{Q}\theta_{j}(s,t)dy(s,t)$ exists.Furthermore

$| \langle\theta_{j},y\rangle|=|\theta_{j}(S, T)y(S, T)-\int_{0}^{T}y(S,t)d\theta_{j}(S,t)$

(3.7)

$- \int_{0}^{s}y(s, T)d\theta_{j}(s, T)+\int_{Q}y(s,t)d\theta_{j}(s,t)|\leq C_{j}\Vert y\Vert_{\infty}$

with

(3.8) $C_{j}=|\theta_{j}(S, T)|+$ Var$(\theta_{j}(S, \cdot), [0, T])+Var(\theta_{j}(\cdot, T), [0,S])+Var(\theta_{j}, Q)<\infty$ .

In Section 4 below, we show that if $F$ and $G$ are elements of $E_{\sigma}(Q)$ then $\mathcal{F}_{a,\beta}F(\cdot),$ $(F*G)_{\alpha}(\cdot)$ ,$\delta F(\cdot|w)$ and $\delta F(y|\cdot)$ exist and are also elements of $E_{\sigma}(Q)$ . Also we examine all relationshipsinvolving exactly two of the three concepts of ”integral transform”, ”convolution product” and“first variation” for functionals in $E_{\sigma}(Q)$ . Furthermore we obtain Parseval‘s and Plancherel‘srelations for functionals in $E_{0}(Q)$ . For related work, see [3, 6, 10, 12, 14, 15, 16, 19, 21] and for adetailed survey of previous work, see [17].

We finish this section by introducing a well-known Yeh-Wiener integration formula for func-tionals $f(\{\vec{\theta},x\rangle)$ :

(3.9) $\int_{C(Q)}f(\{\vec{\theta},x\})m_{Y}(dx)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\vec{u})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$

where $\Vert\vec{u}\Vert^{2}=\sum_{j=1}^{n}u_{j}^{2}$ and $d\vec{u}=du_{1}\cdots du_{n}$ .

\S 4. Integral transform, convolution product and first variation of functionals in $E_{\sigma}(Q)$

In our first theorem, we show that if $F$ is an element of $E_{\sigma}(Q)$ , then the integral transform$\mathcal{F}_{\alpha\beta}F$ of $F$ exists and is an element of $E_{\sigma}(Q)$ .

33

Theorem 4.1. Let $F\in E_{\sigma}(Q)$ be given by (1.7). Then the integml transform $\mathcal{F}_{a\beta}F$ exists,

belongs to $E_{\sigma}(Q)$ and is given by the fomula

(4.1) $\mathcal{F}_{a\beta}F(y)=h(\{\vec{\theta},y\})$

for $y\in K(Q)$ , where

(4.2) $h( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\alpha\vec{u}+\beta\vec{\lambda})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$.

Proof. For each $y\in K(Q)$ , using the Yeh-Wiener integration formula (3.9) we obtain

$\mathcal{F}_{a\beta}F(y)=\int_{C(Q)}f(\alpha\{\tilde{\theta},x\rangle+\beta\langle\vec{\theta},y\})m_{Y}(dx)$

$=(2 \pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\alpha\vec{u}+\beta\langle\vec{\theta},y\})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$

$=h((\vec{\theta},y\})$

where $h$ is given by (4.2). By [8, Theorem 3.15], $h(\tilde{\lambda})$ is an entire function. Moreover by theinequality (1.8) we have

$|h( \tilde{\lambda})|\leq(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}A_{F}\exp\{B_{F}\sum_{j=1}^{n}|\alpha u_{j}+\beta\lambda_{j}|^{1+\sigma}-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$.

But since$|\alpha u_{j}+\beta\lambda_{j}|^{1+\sigma}\leq|2\alpha u_{j}|^{1+\sigma}+|2\beta\lambda_{j}|^{1+\sigma}$ ,

we have

$|h( \vec{\lambda})|\leq A_{\mathcal{F}_{a\beta}F}\exp\{B_{\mathcal{F}_{\alpha\beta}F}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$,

where$A_{\mathcal{F}_{a\beta}F}=(2 \pi)^{-n/2}A_{F}(\int_{\mathbb{R}}\exp\{B_{F}|2\alpha u|^{1+\sigma}-\frac{u^{2}}{2}\}du)^{n}<\infty$

and $B_{\mathcal{F}_{a,\beta}F}=B_{F}(2\beta|)^{1+\sigma}$ . Hence $\mathcal{F}_{a\beta}F\in E_{\sigma}(Q)$. $\square$

In our next theorem we show that the convolution product of functionals from $E_{\sigma}(Q)$ existsand is an element of $E_{\sigma}(Q)$ . We may assume that $F$ and $G$ in Theorem 4.2 below can beexpressed using the same positive integer $n$ . For details see Remark 1.4 of [12].

Theorem 4.2. Let $F,G\in E_{\sigma}(Q)$ be given by (1.7) with corresponding entirefunctions $f$ and$g$. Then the convolution $(F*G)_{\alpha}$ exists, belongs to $E_{\sigma}(Q)$ and is given by the formula

(4.3) $(F*G)_{\alpha}(y)=k(\{\vec{\theta},y\rangle)$

34

for $y\in K(Q)$ , where

(4.4) $k( \vec{\lambda})=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\frac{\vec{\lambda}+\alpha\vec{u}}{\sqrt{2}})g(\frac{\vec{\lambda}-\alpha\vec{u}}{\sqrt{2}})\exp\{-\frac{1}{2}\Vert\tilde{u}\Vert^{2}\}d\vec{u}$ .

Pmof. For each $y\in K(Q)$ , using the Yeh-Wiener integration formula (3.9) we obtain

$(F*G)_{\alpha}(y)= \int_{C(Q)}f(\frac{(\vec{\theta},y\}+\alpha(\vec{\theta},x\}}{\sqrt{2}})g(\frac{\{\vec{\theta},y\}-\alpha\{\vec{\theta},x\rangle}{\sqrt{2}})m_{Y}(dx)$

$=(2 \pi)^{-n/2}\int_{\mathbb{R}^{n}}f(\frac{\{\vec{\theta},y\rangle+\alpha\vec{u}}{\sqrt{2}})g(\frac{\langle\vec{\theta},y\}-\alpha\vec{u}}{\sqrt{2}})\exp\{-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$

$=k(\{\vec{\theta},y\})$

where $k$ is given by (4.4). By [8, Theorem 3.15], $k(\vec{\lambda})$ is an entire function and

$|k( \vec{\lambda})|\leq(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}A_{F}A_{G}\exp\{(B_{F}+B_{G})\sum_{j=1}^{n}(\frac{|\lambda_{j}|+|\alpha u_{j}|}{\sqrt{2}})^{1+\sigma}-\frac{1}{2}\Vert\vec{u}\Vert^{2}\}d\vec{u}$ .

By the same method as in Theorem 4.1, we have

$|k( \vec{\lambda})|\leq A_{(F*G)_{\alpha}}\exp\{B_{(F*G)_{\alpha}}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$ ,

where $B_{(F*G)_{a}}=(B_{F}+B_{G})2^{(1+\sigma)/2}$ and

$A_{(F*G)_{a}}=(2 \pi)^{-n/2}A_{F}A_{G}(\int_{\mathbb{R}}\exp\{(B_{F}+B_{G})(\sqrt{2}|\alpha u|)^{1+\sigma}-\frac{u^{2}}{2}\}du)^{n}<\infty$.

Hence $(F*G)_{\alpha}\in E_{\sigma}(Q)$ . $\square$

In Theorem 4.3 below, we fix $w\in K(Q)$ and consider $\delta F(y|w)$ as a function of $y$, while inTheorem 4.4 below, we fix $y\in K(Q)$ and consider $\delta F(y|w)$ as a function of $w$ .

Theorem 4.3. Let $F\in E_{\sigma}(Q)$ be given by (1.7) and let $w\in K(Q)$ . Then

(4.5) $\delta F(y|w)=p(\langle\vec{\theta},y\})$

for $y\in K(Q)$, where

(4.6) $p( \vec{\lambda})=\sum_{j=1}^{n}\{\theta_{j},w\}f_{j}(\vec{\lambda})$ .

Furthermore, as afunction of$y\in K(Q),$ $\delta F(y|w)$ is an element of $E_{\sigma}(Q)$ .

35

Proof. For $y\in K(Q)$ ,

$\delta F(y|w)=\frac{\partial}{\partial t}f(\{\vec{\theta},y\}+t\{\vec{\theta},w\rangle)|_{t=0}$

$= \sum_{j=1}^{n}(\theta_{j},w\}f_{j}(\{\vec{\theta},y\rangle)=p(\langle\vec{\theta},y\})$

where $p$ is given by (4.6). Since $f(\vec{\lambda})$ is an entire function, $f_{j}(\vec{\lambda})$ and so $p(\vec{\lambda})$ are entire functions.By the Cauchy integral formula we have

$f_{j}( \lambda_{1}, \cdots,\lambda_{j}, \cdots,\lambda_{n})=\frac{1}{2\pi i}\int_{|\zeta-\lambda_{j}|=1}\frac{f(\lambda_{1},\cdots,\zeta,\cdots,\lambda_{n})}{(\zeta-\lambda_{j})^{2}}d\zeta$ .

By the inequality (1.10), for any $\zeta$ with $|\zeta-\lambda_{j}|=1$ , we have

$| \frac{f(\lambda_{1},\cdots,\zeta,\cdots,\lambda_{n})}{(\zeta-\lambda_{j})^{2}}|\leq A_{F}\exp\{B_{F}[|\lambda_{1}|^{1+cr}+\cdots+|\zeta|^{1+(r}+\cdots+|\lambda_{n}|^{1+\sigma}]\}$

$\leq A_{F}\exp\{2^{1+\sigma}B_{F}[\sum_{j=1}^{n}|,t_{j}|^{1+\sigma}+1]\}$ .

Hence

$|f_{j}( \vec{\lambda})|\leq A_{F}\exp\{2^{1+\sigma}B_{F}\}\exp\{2^{1+\sigma}B_{F}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$ ,

and so$|p( \vec{\lambda})|\leq\sum_{j=1}^{n}|(\theta_{j},w\rangle||f_{j}(\vec{\lambda})|\leq A_{\delta F(\cdot|w)}\exp\{B_{\delta F(\cdot|w)}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$

where

$A_{\delta F(\cdot|w)}=A_{F} \exp\{2^{1+\sigma}B_{F}\}\Vert w\Vert_{\infty}\sum_{j=1}^{n}C_{j}<\infty$

with $C_{j}$ given by (3.8) and $B_{\delta F(\cdot|w)}=2^{1+\sigma}B_{F}$ . $\square$

Theorem 4.4. Let $y\in K(Q)$ and let $F\in E_{\sigma}(Q)$ be given by (1.7). Then

(4.7) $\delta F(y|w)=q(\{\vec{\theta},w\rangle)$

for $w\in K(Q)$ , where

(4.8) $q( \vec{\lambda})=\sum_{j=1}^{n}\prime t_{j}f_{j}(\{\vec{\theta},y\rangle)$ .

Furthermore, as afunction of $w,$ $\delta F(y|w)$ is an element of$E_{\sigma}(Q)$ .

36

Pmof. Equations (4.7) and (4.8) are immediate from the first part of the proof of Theorem4.3. Clearly $q(\vec{\lambda})$ is an entire function. Next, using the estimation for $|f_{j}|$ we saw in the proofof Theorem 4.3 above we obtain,

$|q( \vec{\lambda})|\leq\sum_{j=1}^{n}|\lambda_{j}f_{j}(\langle\vec{\theta},y\rangle)|$

$\leq A_{F}\exp\{2^{1+\sigma}B_{F}\}\exp\{2^{1+\sigma}B_{F}\Vert y\Vert_{\infty}^{1+\sigma}(C_{1}^{1+\sigma}+\cdots+C_{n}^{1+\sigma})\}\sum_{j=1}^{n}|\lambda_{j}|$ .

Since $t\leq\exp\{t^{1+\sigma}\}$ for all $t\geq 0$ ,

$\sum_{j=1}^{n}|\lambda_{j}|\leq\exp\{(\sum_{j=1}^{n}|\lambda_{j}|)^{1+\sigma}\}\leq\exp\{n^{1+\sigma}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$

and so

$|q( \vec{\lambda})|\leq A_{\delta F(y|\cdot)}\exp\{B_{\delta F(y|\cdot)}\sum_{j=1}^{n}|\lambda_{j}|^{1+\sigma}\}$

where $B_{\delta F(y|\cdot)}=n^{1+\sigma}$ and

$A_{\delta F(y|\cdot)}=A_{F}\exp\{2^{1+\sigma}B_{F}\}\exp\{2^{1+\sigma}B_{F}\Vert y\Vert_{\infty}^{1+\sigma}(C_{1}^{1+\sigma}+\cdots+C_{n}^{1+\sigma})\}<\infty$ .

Hence, as a function of $w,$ $\delta F(y|w)\in E_{\sigma}(Q)$ . $\square$

Now, we establish all of the various relationships involving exactly two of the three conceptsof“integral transform”, ”convolution product” and ”first variation” for functionals belonging to$E_{\sigma}(Q)$ . The seven distinct relationships, as well as altemative expressions for some of them, aregiven by equations (4.9) through (4.15) below.

In view of Theorem 4.1 through Theorem 4.4 above, all of the functionals that occur in thissection are elements of $E_{\sigma}(Q)$ . For example, let $F$ and $G$ be any functionals in $E_{C\Gamma}(Q)$ . Thenby Theorem 4.2, the functional $(F*G)_{\alpha}$ belongs to $E_{\sigma}(Q)$ , and hence by Theorem 4.1, thefunctional $\mathcal{F}_{\alpha\beta}(F*G)_{a}$ also belongs to $E_{\sigma}(Q)$ . By similar arguments, all of the functionals thatarise in equations (4.9) through (4.19) below, exist and belong to $E_{\sigma}(Q)$ .

Once we have shown the existence theorems (Theorems 4.1 through 4.4 above), the proofsof the Formulas 4.5 through 4.11 below are exactly the same as those in [12]. Hence we juststate the formulas without proofs.

Formula 4.5. The integral transform of the convolution product equals the product of theintegral transforms:

(4.9) $\mathcal{F}_{\alpha\beta}(F*G)_{\alpha}(y)=\mathcal{F}_{\alpha\beta}F(\frac{y}{\sqrt{2}})\mathcal{F}_{\alpha\beta}G(\frac{y}{\sqrt{2}})=\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)\mathcal{F}_{\alpha,\beta/\sqrt{2}}G(y)$

for all $y$ in $K(Q)$ .

37

Formula 4.6. A formula for the convolution product of the integral transform of functionalsfrom $E_{\sigma}(Q)$ :

$(\mathcal{F}_{a\beta}F*\mathcal{F}_{a\beta}G)_{a}(y)$

(4.10)$=(2 \pi)^{-3n/2}\int_{\mathbb{R}^{3n}}f(\alpha\vec{r}+\frac{\beta}{\sqrt{2}}\{\vec{\theta},y\}+\frac{\beta\alpha}{\sqrt{2}}\vec{u})$

$g( \alpha\vec{s}+\frac{\beta}{\sqrt{2}}\{\vec{\theta},y\rangle-\frac{\beta\alpha}{\sqrt{2}}\vec{u})\exp\{-\frac{\Vert\vec{u}\Vert^{2}+\Vert\vec{r}\Vert^{2}+\Vert s\neg|^{2}}{2}\}d\vec{u}d\vec{r}d\vec{s}$

for all $y$ in $K(Q)$ .

Formula 4.7. The integral transform with respect to the first argument of the variationequals $1/\beta$ times the variation of the integral transform:

(4.11) $\mathcal{F}_{\alpha\beta}(\delta F(\cdot|w))(y)=\frac{1}{\beta}\delta \mathcal{F}_{(f}pF(y|w)=\sum_{j=1}^{n}\{\theta_{j},$$w\rangle \mathcal{F}_{tt,\beta}F_{j}(y)$

for all $y$ and $w$ in $K(Q)$ .

Formula 4.8. The transform with respect to the second argument of the variation equals $\beta$

times the variation of the functional:

(4.12) $\mathcal{F}_{\alpha\beta}(\delta F(y|\cdot))(w)=\beta\delta F(y|w)$

for all $y$ and $w$ in $K(Q)$ .

Formula4.9. A formula for the first variation of the convolution product of functionalsfrom $E_{\sigma}(Q)$ :

(4.13) $\delta(F*G)_{\alpha}(y|w)=\sum_{j=1}^{n}\frac{\langle\theta_{j},w\}}{\sqrt{2}}[(F_{j}*G)_{\alpha}(y)+(F*G_{j})_{\alpha}(y)]$

for all $y$ and $w$ in $K(Q)$ .

Formula 4.10. A formula for the convolution product, with respect to the first argument of

the variation, of the variation of functionals from $E_{\sigma}(Q)$ :

(4.14) $( \delta F(\cdot|w)*\delta G(\cdot|w))_{a}(y)=\sum_{j=1}^{n}\sum_{l=1}^{n}\{\theta_{j},w\rangle\{\theta_{l},w\rangle(F_{j}*G_{l})_{a}(y)$

for all $y$ and $w$ in $K(Q)$ .

Formula 4.11. A formula for the convolution product, with respect to the second argumentof the variation, of the variation of functionals from $E_{\sigma}(Q)$ :

(4.15) $( \delta F(y|\cdot)*\delta G(y|\cdot))_{a}(w)=\frac{1}{2}\delta F(y|w)\delta G(y|w)-\frac{\alpha^{2}}{2}\sum_{j=1}^{n}F_{j}(y)G_{j}(y)$

for all $y$ and $w$ in $K(Q)$ .

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Finally, letting $G=F$ in equations (4.9), (4.13), (4.14) and (4.15) above, yields the formulas

(4.16) $\mathcal{F}_{\alpha\beta}(F*F)_{\alpha}(y)=[\mathcal{F}_{\alpha\beta/\sqrt{2}}F(y)]^{2}$ ,

(4.17) $\delta(F*F)_{a}(y|w)=\sqrt{2}\sum_{j=1}^{n}\langle\theta_{j},w\rangle(F*F_{j})_{\alpha}(y)$ ,

(4.18) $( \delta F(\cdot|w)*\delta F(\cdot|w))_{\alpha}(y)=\sum_{j=1}^{n}\sum_{l=1}^{n}\{\theta_{j},w\}\{\theta_{l},w\}(F_{j}*F_{l})_{\alpha}(y)$ ,

and

(4.19) $( \delta F(y|\cdot)*\delta F(y|\cdot))_{\alpha}(w)=\frac{1}{2}[\delta F(y|w)]^{2}-\frac{\alpha^{2}}{2}\sum_{j=1}^{n}[F_{j}(y)]^{2}$

for all $y$ and $w$ in $K(Q)$ .Furthermore we can obtain the following Parseval’s and Plancherel’s relation.Let $H_{0}=H_{0}(Q)$ be the space of real-valued functions $f$ on $Q$ which are absolutely continu-

ous and whose derivative $Df$ is in $L_{2}(Q)$ . The inner product on $H_{0}$ is given by

$\langle f,g\rangle=\int_{Q}(Df)(s)(Dg)(s)ds$ .

Then $H_{0}$ is a real separable infinite dimensional Hilbert space. Let $B_{0}=B_{0}(Q)$ be the Yeh-Wiener space $C(Q)$ and equip $B_{0}$ with the $\sup$ norm. Then $(H_{0},B_{0},m_{Y})$ is an example of anabstract Wiener space.

We restrict our attention, in this subsection, to the space $E_{0}(Q)$ rather than $E_{\sigma}(Q)$ . Now itis well known, see for example [6,15], that for all $F\in E_{0}(Q)$ , all $y\in K(Q)$ and all complexnumbers $a,b$ and $c$ ,

(4.20) $\int_{C(Q)}\int_{C(Q)}F(ax+by+cw)m_{Y}(dx)m_{Y}(dy)=\int_{C(Q)^{F(\sqrt{a^{2}+b^{2}}}}z+cw)m_{Y}(dz)$

and that

(4.21) $\mathcal{F}_{\alpha,\beta}(\mathcal{F}_{(x’\beta’}F)(y)=F(y)=\mathcal{F}_{\alpha’\beta’}(\mathcal{F}_{a,\beta}F)(y)$

provided $\beta\beta’=1$ and $\alpha^{2}+(\beta\alpha’)^{2}=0$ .

Theorem 4.12. Let $F,G\in E_{0}(Q)$ and let $\alpha’$ be a complex number such that $\alpha^{2}+(\beta\alpha’)^{2}=0$ .Then Parseval’s relation

(4.22) $\int_{C(Q)}\mathcal{F}_{a_{:}\beta}F(\frac{\alpha’y}{\sqrt{2}})\mathcal{F}_{\alpha\beta}G(\frac{\alpha’y}{\sqrt{2}})$ $my$ $(dy)= \int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})G(-\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$

holds. In particular, $\iota f\beta=i$, we have

(4.23) $\int_{C(Q)}\mathcal{F}_{\alpha,i}F(\frac{\alpha y}{\sqrt{2}})G(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)=\int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{\alpha,i}G(\frac{\alpha y}{\sqrt{2}})$$my$ $(dy)$ .

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Moreover, formula (4.23) induces Plancherel’s relation ofthe form

(4.24) $\int_{C(Q)}|\mathcal{F}_{a,i}F(\frac{\alpha y}{\sqrt{2}})|^{2}$ $my$ $(dy)= \int_{C(Q)}|F(\frac{\alpha y}{\sqrt{2}})|^{2}$ $my$ $(dy)$ .

Proof. From Formula 4.5 and Definition 3.1, it follows that the left hand side of (4.22) isequal to

$\int_{C(Q)}\mathcal{F}_{a\beta}(F*G)_{a}(\alpha’y)m_{Y}(dy)$

$= \int_{C(Q)}\int_{C(Q)}(F*G)_{a}(\alpha x+\beta\alpha’y)m_{Y}(dx)m_{Y}(dy)$ .

But by (4.20) and the fact that $\alpha^{2}+(\beta\alpha’)^{2}=0$ , the last integral is equal to $(F*G)_{\alpha}(O)$ , which isequal to the right hand side of (4.22).

From (4.21) we know that $\mathcal{F}_{a,i}(\mathcal{F}_{a,-i}G)(y)=G(y)$ and so we have

$\int_{C(Q)}\mathcal{F}_{a,i}F(\frac{\alpha y}{\sqrt{2}})G(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$

$= \int_{C(Q)}\mathcal{F}_{x,i}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{\alpha,i}(\mathcal{F}_{\alpha,-i}G)(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$

$= \int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{a,-i}G(-\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$ ,

where the second equality is obtained by (4.22). But it is easy to see that $\mathcal{F}_{a,-i}G(-\alpha y/\sqrt{2})=$

$\mathcal{F}_{a,i}G(\alpha y/\sqrt{2})$ and this completes the proof of (4.23).

Finally, since $\mathcal{F}_{\alpha,i}F(\alpha y/\sqrt{2})=\mathcal{F}_{\overline{a},-\overline{\iota a}/a}\overline{F}(\alpha y/\sqrt{2})$ , by (4.23) we have

$\int_{C(Q)}|\mathcal{F}_{\alpha,i}F(\frac{\alpha y}{\sqrt{2}})|^{2}m_{Y}(dy)=\int_{C(Q)}F(\frac{\alpha y}{\sqrt{2}})\mathcal{F}_{\alpha,i}\mathcal{F}_{\overline{\alpha},-\overline{\iota\alpha}/\alpha}\overline{F}(\frac{\alpha y}{\sqrt{2}})m_{Y}(dy)$ .

But by (4.20), it is easy to see that $\mathcal{F}_{a,i}\mathcal{F}_{\overline{a},-\overline{\iota a}/\alpha}\overline{F}(\alpha y/\sqrt{2})=F(\alpha y/\sqrt{2})$ and this completes theproof of (4.24). $\square$

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